Question #affe9

1 Answer
Jan 3, 2018

The formula is #"C"_ 6"H"_ {12}"O"_ 6#.

Explanation:

You are supposed to convert mass to moles using atomic and molecular weights.

Start with one mole of the carbohydrate. That is given as #180.2"g"#. Then the amount of carbon is

#({40%"C"}/{100%})×({180.2"g"}/"mol")=72.1"g C"#

Since to the nearest tenth, #100-53.3-6.7=40.0#, I assume that the #40%# carbon has one significant zero after the decimal point (#40.0%#). Hence the use of three significant digits.

And we then convert the carbon to moles by dividing by the atomic weight. Most periodic tables give this as #12.01# so:

#(72.1"g C")({1"mol"}/{12.01"g"})=6.00"mol C"#

So one mole of the carbohydrate has six moles of carbon. That means the molecular formula has #"C"_6#.

For oxygen we do the same thing, but with the oxygen numbers:

#({53.3%"O"}/{100%})×({180.2"g"}/"mol")=96.0"g O"#

#(96.0"g O")({1"mol"}/{15.999"g"})=6.00"mol O"#

So the formula will contain six oxygen atoms. Now we have #"C"_ 6"H" _{??}"O" _6#, as formulas for organic compounds are usually written with carbon and hydrogen first.

I will let you do the calculations for hydrogen. You should find the formula will contain #"H"_ {12}#. Thus #"C"_ 6"H"_ {12}"O"_ 6#.